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Make $x$ the subject of the equation:

$4^{ax} = b * 8^x$

I have tried to substitute the powers of 2 into the equation and have so far got

$2^{2ax} = b * 2^{3x}$

however I am now stuck. If you can, can you please show me ways to get $x$ on its own? Also I have just started A levels and know nothing about logarithms

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4 Answers

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Noting that $4^{ax}=2^{2ax}$ and $8^x=2^{3x}$, using the laws of logs $\ln(ab)=\ln a+\ln b$ and $\ln(a^b) = b\ln a$, we get

\begin{align*} &4^{ax} = b \times 8^x\\ \implies &2^{2ax}=2^{3x}\,b\\ \implies &\ln(2^{2ax})=\ln(2^{3x}\,b)\\ \implies & 2ax\ln2=3x\ln2+\ln b\\ \implies &x\ln 2(2a-3)=\ln b\\ \therefore~~&\boxed{\,x=\frac{\ln b}{(2a-3)\ln2}} \end{align*}

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Hint:

$\log(a\cdot b^x) = \log(a)+x\log(b)$

So, supposing $4^{ax}=b\cdot 8^x$, by applying $\log_2$ to each side we have $\log_2(4^{ax})=\log_2(b\cdot 8^x)$ which simplifies to...

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Observing that $4=2^2$ and $8=2^3$ gives $$(2^2)^{ax}=(2^3)^x\cdot b,$$ or equivalently $$2^{2ax}=2^{3x}\cdot b.$$ Now, taking logarithms on both sides gives $$2ax\log(2)=3x\log(2)+\log(b),$$ then $$2ax\log(2)-3x\log(2)=\log(b).$$ factorising $x\log(2)$ out $$x\log(2)(2a-3)=\log(b).$$ Thus $$x=\frac{\log(b)}{\log(2)(2a-3)}.$$

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So ... $$2^{2ax-3x}=b$$ i.e. $$(2a-3)x=\log_2 b$$ i.e. $$x=\frac{\log_2 b}{2a-3}$$ I don't believe there is any way you can avoid using logarithms.

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